The following solutions were prepared by… - NEET UG 2021 Chemistry PYQ

NEET UG 2021 — Chemistry Physical Chemistry

2021mcqmedium

The following solutions were prepared by dissolving $10g$ of glucose $({C}_{6}{H}_{12}{O}_{6})$ in $250\mathrm{mL}$ of water $({P}_{1}),10g$ of urea $({\mathrm{CH}}_{4}{N}_{2}O)$ in $250\mathrm{mL}$ of water $({P}_{2})$ and $10g$ of sucrose $({C}_{12}{H}_{22}{O}_{11})$ in $250\mathrm{mL}$ of water $({P}_{3}).$ The right option for the decreasing order of osmotic pressure of these solutions is:

Official previous-year question

Held on 30 Apr 2021 · Verified 9 Jul 2026.

Options

A
${P}_{2}>{P}_{1}>{P}_{3}$
B
${P}_{1}>{P}_{2}>{P}_{3}$
C
${P}_{2}>{P}_{3}>{P}_{1}$
D
${P}_{3}>{P}_{1}>{P}_{2}$

Did you get this right?

Solution

Osmotic pressure $(\pi )=\mathrm{CRT}$

At same temperature, $\pi \propto C$

Solution: $(1)$ ${C}_{1}=\frac{10}{180}=\frac{10}{180}\times 4$

${C}_{1}=\frac{250}{1000}=\frac{40}{180}=\frac{2}{9}$

${C}_{1}=0.222M$

Solution: $(2)$ ${C}_{2}=\frac{10}{\frac{60}{\frac{250}{1000}}}=\frac{1}{6}\times 4=0.667M$

Solution: $(3)$ ${C}_{3}=\frac{10}{342}\times \frac{1000}{250}=0.1169M$

$\because {C}_{2}>{C}_{1}>{C}_{3}\therefore$ Hence, order: ${P}_{2}>{P}_{1}>{P}_{3}$

Your note

NEET UG Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every NEET UG Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

Options

Did you get this right?

Solution

Your note

NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry